Timeline for Dense sets in the space of continuous functions
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2011 at 22:16 | comment | added | George Lowther | @guykatriel: By the Hahn-Jordan decomposition you have a Borel set $A$ with $\mu^+(E)=\mu(E\cap A)$ and $\mu^-(E)=-\mu(E\setminus A)$ for all $E$. If $\mu^+(S)\le\mu^-(S)$ for all Borel $S\subseteq V$ then that would imply $$0\le\mu^+(V)=\mu^+(V\cap A)\le\mu^-(V\cap A)=0.$$ | |
| Sep 21, 2011 at 15:02 | vote | accept | user17970 | ||
| Sep 21, 2011 at 15:02 | vote | accept | user17970 | ||
| Sep 21, 2011 at 15:02 | |||||
| Sep 21, 2011 at 7:12 | comment | added | user17970 | .. I mean: to posit that the restriction of $\mu$ to any open set is not a positive measure and not a negative measure. | |
| Sep 21, 2011 at 7:02 | comment | added | user17970 | Can you clarify one point: why would $\mu^+(S)\leq \mu^-(S)$ for all Borel $S\subset V$ imply that $\mu^+(V)=0$? I think you have to posit in advance that the measure $\mu$ is not positive and not negative on any open set. | |
| Sep 20, 2011 at 22:33 | history | edited | George Lowther | CC BY-SA 3.0 |
deleted 27 characters in body; edited body; added 2 characters in body
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| Sep 20, 2011 at 22:25 | history | answered | George Lowther | CC BY-SA 3.0 |